In: Statistics and Probability
A sociologist is studying the prevalence of crime in one major city. In a sample of 150 randomly selected residents, 54 say that they have been victimized by a criminal. Based on this sample, construct a 99% confidence interval for the proportion of all residents in this city who have been victimized by a criminal. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. What is the lower limit of the confidence interval? What is the upper limit of the confidence interval?
Solution :
n = 150
x = 54
= x / n = 54 / 150 = 0.360
1 - = 1 - 0.36 = 0.640
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 = 2.576
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 2.576 * (((0.360 * 0.640) / 150)
= 0.10
A 99 % confidence interval for population proportion p is ,
- E < P < + E
0.360 - 0.10 < p < 0.360 + 0.10
0.260< p < 0.460
The lower limit = 0.26
The upper limit = 0.46